Mathematics · Parabola

JEE Main 2025 — 22 January, Evening Shift — Question 3

Let $P(4,4 \sqrt{3})$ be a point on the parabola $y^{2}=4 \mathrm{ax}$ and PQ be a focal chord of the parabola. If $M$ and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:

Option A:
$\frac{263 \sqrt{3}}{8}$
Option B:
$17 \sqrt{3}$
Option C:
$\frac{343 \sqrt{3}}{8}$
Correct
Option D:
$\frac{34 \sqrt{3}}{3}$

Answer: C

Step-by-step solution

$(4,4 \sqrt{3})$ lies on $y^{2}=4 \mathrm{ax}$

$\Rightarrow 48=4 \mathrm{a} .4$ $4 a=12$

$\Rightarrow y^{2}=12 x$ is equation of parabola

Now, parameter of P is $\mathrm{t}_{1}=\frac{2}{\sqrt{3}} \Rightarrow$ Parameters of Q is

$\mathrm{t}_{2}=-\frac{\sqrt{3}}{2} \Rightarrow \mathrm{Q}\left(\frac{9}{4},-3 \sqrt{3}\right)$

Area of trapezium PQNM

$=\frac{1}{2} \mathrm{MN} .(\mathrm{PM}+\mathrm{QN})$

$=\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PS}+\mathrm{QS})$

$=\frac{1}{2} \mathrm{MN} . \mathrm{PQ}$

$=\frac{1}{2} 7 \sqrt{3} \cdot \frac{49}{4}=(343) \frac{\sqrt{3}}{8}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Mathematics
Chapter: Parabola
Topic: Chords connected with a Parabola